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(A) Step $1$: Analyze the circuit structure. The $6 \Omega$ and $2 \Omega$ resistors are connected in series with each other in the lower branch. Let

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$6$

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(A) Step $1$: Analyze the circuit structure. The $6 \Omega$ and $2 \Omega$ resistors are connected in series with each other in the lower branch. Let their equivalent resistance be $R_1 = 6 \Omega + 2 \Omega = 8 \Omega$. However,looking at the circuit diagram,the $6 \Omega$ and $2 \Omega$ resistors are in series,and this combination is in parallel with the $1.5 \Omega$ resistor. Let's re-evaluate: The $6 \Omega$ and $2 \Omega$ are in series,$R_s = 6 + 2 = 8 \Omega$. This $8 \Omega$ is in parallel with $1.5 \Omega$. $R_p = (8 	imes 1.5) / (8 + 1.5) = 12 / 9.5 \approx 1.26 \Omega$. This does not match the options. Let's re-examine the diagram: The $6 \Omega$ and $2 \Omega$ are in series,and the $1.5 \Omega$ is in parallel with the $6 \Omega$ resistor? No,the diagram shows $1.5 \Omega$ and $6 \Omega$ in series,and $2 \Omega$ in parallel to that branch. Let $R_s = 1.5 + 6 = 7.5 \Omega$. Then $R_p = (7.5 	imes 2) / (7.5 + 2) = 15 / 9.5 \approx 1.57 \Omega$. Finally,this is in parallel with $3 \Omega$. $R_{eq} = (1.57 	imes 3) / (1.57 + 3) \approx 1.04 \Omega$. Still not matching. Let's assume the standard interpretation: $6 \Omega$ and $2 \Omega$ are in parallel,$R' = (6 	imes 2) / (6 + 2) = 1.5 \Omega$. This $1.5 \Omega$ is in series with the $1.5 \Omega$ resistor,$R'' = 1.5 + 1.5 = 3 \Omega$. This $3 \Omega$ is in parallel with the top $3 \Omega$ resistor,$R_{eq} = (3 	imes 3) / (3 + 3) = 1.5 \Omega$. Step $2$: Calculate the total current $I$ using Ohm's Law: $I = V / R_{eq} = 9 	ext{ V} / 1.5 \Omega = 6 	ext{ A}$.

The total current supplied to the given circuit by the battery is . . . . . . . (in $A$)

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